Suppose $\omega$ is a real closed $(1,1)$ form on a compact Kähler manifold. If we have a real $d$-closed two form $\sigma$ such that $[\sigma]=[\omega] \in H^2(M)$, can we claim that this two form $\sigma$ is also a $(1,1)$ form?
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$\begingroup$ This seems too sloppily asked to be answered. First, shouldn't the last $\omega$ be a $\sigma$? Second, doesn't the very notation "$[\sigma]\in H^{1,1}(M)$" already assume that $\sigma$ is a $\overline\partial$-closed $(1,1)$ form, hence an affirmative answer to your question? $\endgroup$– Francois ZieglerJul 12, 2013 at 1:44
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$\begingroup$ sorry. My fault. I modified the question and I hope the question now is clear. $\endgroup$– KevinJul 12, 2013 at 1:50
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$\begingroup$ $\sigma = \omega +d\alpha$ for $\alpha$ a suitably chosen 1-form is a counter example. $\endgroup$– Peter MichorJul 12, 2013 at 5:31
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$\begingroup$ @PeterMichor How to choose this $\alpha$? It seems to me not so obvious. $\endgroup$– KevinJul 12, 2013 at 5:44
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1 Answer
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I think not. Note that replacing $\sigma$ by $\sigma - \omega$ reduces us to the case $\omega=0$. Your question in that case is whether every real $d$-exact 2-form is a $(1,1)$-form.
Now unless I'm mistaken, on projective space $\mathbf{CP}^2$ with homogeneous coordinates $(a:b:c)$, we get a real $d$-exact 2-form with components of all types ($(2,0)$, $(1,1)$, $(0,2)$) by putting $$ \sigma =d\left(\frac{A}{A+B+C} d\left(\frac{B}{A+B+C}\right)\right) =\frac{AdB\wedge dC+BdC\wedge dA+CdA\wedge dB}{(A+B+C)^3} $$ where $(A,B,C)=(|a|^2, |b|^2, |c|^2)$.
answered Jul 12, 2013 at 6:02